Draw a plot showing the variation of (i) electric field (E) and (ii) electric potential (V) with distance r due to a point charge Q.    from Physics Class 12 CBSE Year 2012 Free Solved Previous Year Papers

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Physics

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CBSE Class 12

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CBSE Physics 2012 Exam Questions

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11.

A ray of light, incident on an equilateral glass prism ( bold italic mu subscript bold italic g equals square root of 3) moves parallel to the base line of the prism inside it. Find the angle of incidence for this ray.


Since the prism is equilateral in shape, 



Angle of prism, A = 60o

Angle of refraction, r = A/2 = 30o

Now, using Snell’s law, 

mu subscript g over mu subscript i equals fraction numerator sin space i over denominator sin space r end fraction 


fraction numerator square root of 3 over denominator 1 end fraction equals fraction numerator sin space i over denominator sin space 30 degree end fraction 

rightwards double arrow straight space sin straight space straight i straight space equals straight space fraction numerator square root of 3 over denominator 2 end fraction
i. e. comma space A n g l e space o f space i n c i d e n c e comma space i space equals space 60 degree

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12.

Define self-inductance of a coil. Show that magnetic energy required to build up the current I in a coil of self-inductance L is given by 1 half LI squared


 Self-inductance of a coil is numerically equal to the magnetic flux linked with the coil when the current through coil is one Ampere.

Mathematically, it is given by, 

                                               straight phi straight space equals straight space LI

where, L is the constant of proportionality and is called the self-inductance.

Energy stored in an inductor:

Consider a source of emf connected to an inductor L.

With increase in current, the opposing induced emf is given by, straight e straight space equals straight space minus straight L di over dt
If the source of emf sends a current i through the inductor for a small time dt, then the amount of work done by the source, 

dW straight space equals vertical line straight e vertical line straight i straight space dt straight space equals straight space Li    di over dt dt straight space equals straight space Li straight space di

Hence, the total amount of work done by source of emf when the current increases from its initial values (i = 0) to its final value (I) is given by,

Error converting from MathML to accessible text.

This work done gets stored in the inductor in the form of energy.

Therefore comma space Energy space stored space in space the space magnetic space inductor comma space straight U space equals 1 half LI squared

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13.

State de-Broglie hypothesis.


According to hypothesis of de Broglie "The atomic particles of matter moving with a given velocity, can display the wave like properties.

i.e., straight lambda straight space equals straight space straight h over mv


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14.

A rectangular loop of wire of size 4 cm × 10 cm carries a steady current of 2A. A straight long wire carrying 5A current is kept near the loop as shown. 


If the loop and the wire are coplanar, find

(i) the torque acting on the loop and

(ii) the magnitude and direction of the force on the loop due to the current carrying wire.


i) Torque on the current carrying conductor is given by, 

straight tau straight space equals straight space straight M with rightwards harpoon with barb upwards on top straight space cross times straight B with rightwards harpoon with barb upwards on top straight space equals straight space MB straight space sin straight space straight theta

In this case, M and B have the same direction. Therefore,  = 0o.

So Torque,  = 0.

ii) The magnetic forces are equal and opposite on the parallel current carrying wires.

Therefore, force acting on the loop is given by, 

Error converting from MathML to accessible text.

The force is attractive in nature or is towards the conductor.

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15.

A metallic rod of ‘L’ length is rotated with angular frequency of ‘ straight omega ’ with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius L, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field B parallel to the axis is present everywhere. Deduce the expression for the emf between the centre and the metallic ring.


Emf is induced is given by, epsilon space equals fraction numerator d phi subscript B over denominator d t end fraction equals straight d over dt left parenthesis BA right parenthesis straight space equals straight space straight B straight space dA over dt
where, dA over dt comma spaceis the rate of change of area of the loop formed by the sector OPQ.

At any instant of time t, let  be the angle between the rod and the radius of the circle at P.

Area of the sector OPQ = straight R squared cross times fraction numerator straight theta over denominator 2 straight pi end fraction equals 1 half straight R squared straight theta ;R is the radius of the circle.

Therefore, emf induced is, Error converting from MathML to accessible text. 

                                     straight epsilon equals straight space 1 half BR squared dθ over dt equals BωR squared over 2

 

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16.

Distinguish between ‘Analog and Digital signals’.


Difference between analog and digital signals is:

Analog signal

Digital signal

They are continuous signals i.e., continuous variation of voltage or current.

These are the signals which can take only discrete values. In digital signal, high means 1 and low means 0.

E.g. Sound of a human

E.g., Temperature



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17.

Using Kirchhoff’s rules determine the value of unknown resistance R in the circuit so that no current flows through 4  resistance. Also find the potential difference between A and D.


Applying Kirchhoff's loop rule for loop ABEFA,

              –9 + 6 + 4 × 0 + 2I = 0

                                          I = 1.5 A                                           … (i)

For loop BCDEB,

3 + IR + 4 × 0 – 6 = 0

Therefore,        IR = 3

Putting the value of I from equation (i), we have

3 over 2 cross times R space equals space 3
rightwards double arrow space space straight space straight R straight space equals straight space 2 straight capital omega

Potential difference between A and D through path ABCD,

9 straight space – straight space 3 straight space – straight space IR straight space equals straight space straight V subscript AD straight space end subscript
rightwards double arrow straight space 9 straight space – straight space 3 straight space – straight space 3 over 2 cross times straight space 2 straight space equals straight space straight V subscript AD
rightwards double arrow space space space space space space space space space space space space space space space space space straight V subscript AD straight space equals straight space 3 straight space straight V

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18.

Under what condition does a biconvex lens of glass having a certain refractive index act as a plane glass sheet when immersed in a liquid?


When immersed in a liquid, a bi-convex lens of glass will behave as a plane glass, 

when, mu subscript L space equals space mu subscript gmu subscript L is the refractive index of liquid and  is the refractive index of glass. 
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19.

Draw a plot showing the variation of (i) electric field (E) and (ii) electric potential (V) with distance r due to a point charge Q.   


For a point charge Q,

Electric potential is inversely proportional to r and Electric field is inversely proportional to r2.

Electric Potential

Electric field

 V space equals space fraction numerator Q over denominator 4 pi epsilon subscript o r end fraction semicolon space i. e. comma space V proportional to 1 over r  E equals fraction numerator Q over denominator 4 pi epsilon subscript o r squared end fraction semicolon space i. e. comma space E proportional to 1 over r squared


The graph below shows us the variation of E and V with distance ‘r’.

 

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20.

The figure shows a series LCR circuit with L = 5.0 H, C = 80 mF, R = 40 W connected to a variable frequency 240V source. Calculate. 


(i) The angular frequency of the source which drives the circuit at resonance.

(ii) The current at the resonating frequency.

(iii) The rms potential drop across the capacitor at resonance.


Given a series LCR circuit,

i) At resonance, angular frequency is given by, 

        bold italic omega subscript bold italic r space equals space fraction numerator 1 over denominator square root of bold italic L bold italic C end root end fraction equals fraction numerator 1 over denominator square root of 5 cross times 80 cross times 10 to the power of negative 6 end exponent end root end fraction

space space space space space equals space 50 space rad divided by sec

ii) Current at resonating frequency is given by, 

        straight I subscript rms straight space equals straight space straight V subscript rms over straight R equals straight space 240 over 40 equals straight space 6 straight space straight A
iii) RMS potential drop across the capacitor at resonance is given by, 

V subscript r m s end subscript italic space equals I subscript r m s end subscript cross times space chi subscript c

italic space italic space italic space italic space italic space italic space italic space equals space 6 cross times fraction numerator italic 1 over denominator italic 50 italic cross times italic 80 italic cross times italic 10 to the power of italic minus italic 6 end exponent end fraction

space space space space space space space equals space fraction numerator 6 cross times 10 to the power of 6 over denominator 4 cross times 10 cubed end fraction italic space

space space space space space space space equals space 1500 space straight V

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